Problem 31 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$1-\sin x=\cos 2 x$$

Step-by-Step Solution

Answer

Solutions: in radians, \( x = n\pi, \frac{\pi}{6} + 2n\pi, \frac{5\pi}{6} + 2n\pi \); in degrees, \( x = 0^\circ, 180^\circ, 30^\circ, 150^\circ \).

Step 1: Use a Trigonometric Identity

We start by recognizing that the cosine double angle formula can be useful here. We know that \( \cos 2x = 1 - 2\sin^2 x \). Therefore, we can rewrite the original equation \( 1 - \sin x = \cos 2x \) as: \( 1 - \sin x = 1 - 2\sin^2 x \).

Step 2: Simplify the Equation

Cancel the 1s from both sides, leaving us with \( -\sin x = -2\sin^2 x \). Dividing the entire equation by \(-1\), we get \( \sin x = 2\sin^2 x \).

Step 3: Factor the Equation

Rearrange the equation to \( 2\sin^2 x - \sin x = 0 \). Factor out \( \sin x \) to have \( \sin x (2\sin x - 1) = 0 \).

Step 4: Solve for Each Factor

We have two solutions from the factored equation: \( \sin x = 0 \) and \( 2\sin x - 1 = 0 \).- For \( \sin x = 0 \), the solutions are \( x = n\pi \), where \( n \) is an integer.- For \( 2\sin x - 1 = 0 \), solve for \( \sin x = \frac{1}{2} \). This gives solutions \( x = \frac{\pi}{6} + 2n\pi \) or \( x = \frac{5\pi}{6} + 2n\pi \), where \( n \) is any integer.

Step 5: Find Exact Solutions in Degrees

Convert the radian measures to degrees. Recall \( x = \frac{\pi}{6} \) is \( 30^\circ \) and \( x = \frac{5\pi}{6} \) is \( 150^\circ \). Therefore, the solutions in degrees for the exact values are \( x = 0^\circ, 180^\circ, 30^\circ, 150^\circ \) for their respective cycles.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in solving equations involving trigonometric functions. These identities express relationships between different trigonometric ratios, like sine and cosine. One such identity is the cosine double angle formula, which is particularly useful for transforming equations. The formula is given by \( \cos 2x = 1 - 2\sin^2 x \), which helps in rewriting equations to a simpler form.
In solving an equation like \( 1 - \sin x = \cos 2x \), applying the cosine double angle identity allows us to remove \( \cos 2x \) and rewrite the equation entirely in terms of \( \sin x \). This is extremely helpful because it reduces the complexity of the problem and paves the way for easier manipulation and solving.

Radian and Degree Conversion

When working with trigonometric functions, understanding the conversion between radians and degrees is crucial. The reason lies in the dual usage of angles in mathematics: radians are often used in calculus and theoretical settings, whereas degrees are more common in practical and everyday applications.
The conversion is straightforward with the formula \( x^\circ = x \times \left( \frac{180}{\pi} \right) \), where \( x \) in degrees is transformed to radians, and vice versa. For example, converting \( \frac{\pi}{6} \) radians to degrees involves multiplying by \( \frac{180}{\pi} \), resulting in \( 30^\circ \).
Understanding this conversion is essential when you are required to express your solution in both radians and degrees, ensuring that any applications of the solutions are accurate in whichever unit is needed.

Trigonometric Factoring

Factoring is a powerful method to solve trigonometric equations, just as it is with algebraic equations. Once you simplify an equation using a trigonometric identity, factoring can dissect the expression into manageable pieces. For example, consider the equation \( 2\sin^2 x - \sin x = 0 \).
To factor this, identify the common factor, which in this case is \( \sin x \). By factoring out \( \sin x \), the equation becomes \( \sin x (2\sin x - 1) = 0 \). This setup allows us to solve simpler equations like \( \sin x = 0 \) and \( 2\sin x - 1 = 0 \) separately.
By factoring, the original equation is effectively broken down into smaller chunks, making it possible to apply easier methods to find multiple solutions.

Solution Sets for Trigonometric Functions

Solving trigonometric equations often results in multiple solutions due to the periodic nature of sine and cosine functions. Once the equation is factored, like in our case \( \sin x = 0 \) and \( 2\sin x - 1 = 0 \), solution sets can be found by considering the periodicity.
For \( \sin x = 0 \), solutions such as \( x = n\pi \) signify every half rotation (\( \pi \) radians) results in a duplicating sine value, where \( n \) is an integer that captures all these occurrences.
For \( 2\sin x - 1 = 0 \), solving gives \( \sin x = \frac{1}{2} \), providing solutions like \( x = \frac{\pi}{6} + 2n\pi \) and \( x = \frac{5\pi}{6} + 2n\pi \). Here, the \( 2n\pi \) component reflects the integer multiples of a full rotation, allowing us to capture all angles fulfilling the condition.
Thus, understanding the structure of these solution sets ensures no potential angle is overlooked, offering comprehensive results in both degrees and radians.